Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $x = \dfrac{2a - 12}{-3a^2 + 27a - 54} \times \dfrac{-a^3 - 6a^2 + 27a}{a^2 + 3a} $
First factor out any common factors. $x = \dfrac{2(a - 6)}{-3(a^2 - 9a + 18)} \times \dfrac{-a(a^2 + 6a - 27)}{a(a + 3)} $ Then factor the quadratic expressions. $x = \dfrac {2(a - 6)} {-3(a - 3)(a - 6)} \times \dfrac {-a(a - 3)(a + 9)} {a(a + 3)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {2(a - 6) \times -a(a - 3)(a + 9) } { -3(a - 3)(a - 6) \times a(a + 3)} $ $x = \dfrac {-2a(a - 3)(a + 9)(a - 6)} {-3a(a - 3)(a - 6)(a + 3)} $ Notice that $(a - 3)$ and $(a - 6)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {-2a\cancel{(a - 3)}(a + 9)(a - 6)} {-3a\cancel{(a - 3)}(a - 6)(a + 3)} $ We are dividing by $a - 3$ , so $a - 3 \neq 0$ Therefore, $a \neq 3$ $x = \dfrac {-2a\cancel{(a - 3)}(a + 9)\cancel{(a - 6)}} {-3a\cancel{(a - 3)}\cancel{(a - 6)}(a + 3)} $ We are dividing by $a - 6$ , so $a - 6 \neq 0$ Therefore, $a \neq 6$ $x = \dfrac {-2a(a + 9)} {-3a(a + 3)} $ $ x = \dfrac{2(a + 9)}{3(a + 3)}; a \neq 3; a \neq 6 $